Question

Derive the following derivative formulas given that \(\frac{d}{d x}(\cosh x)=\sinh x\)and\(\frac{d}{d x}(\sinh x)=\cosh x\). $$\frac{d}{d x}(\operatorname{coth} x)=-\operatorname{csch}^{2} x$$

Short answer

Question: Derive the derivative formula for hyperbolic cotangent (coth x) using the given derivatives for hyperbolic cosine and hyperbolic sine. Solution: The derivative formula for hyperbolic cotangent (coth x) is derived as follows: 1. Express coth x as a ratio between hyperbolic cosine and hyperbolic sine: $$\operatorname{coth} x =\frac{\cosh x}{\sinh x}$$ 2. Apply the quotient rule to find the derivative: $$\frac{d}{d x}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$$ 3. Substitute f(x) and g(x) with their respective derivatives: $$\frac{d}{d x}(\operatorname{coth} x) = \frac{\sinh x \cdot \sinh x - \cosh x \cdot \cosh x}{(\sinh x)^2}$$ 4. Simplify the expression: $$\frac{d}{d x}(\operatorname{coth} x) = \frac{\sinh^2 x - \cosh^2 x}{\sinh^2 x}$$ 5. Use the hyperbolic identity \(\cosh^2 x - \sinh^2 x = 1\): $$\frac{d}{d x}(\operatorname{coth} x) = \frac{-1}{\sinh^2 x}$$ 6. Express the final result in terms of csch^2 x: $$\frac{d}{d x}(\operatorname{coth} x) = - \operatorname{csch}^2 x$$

Step-by-step solution

Define coth x

We know that the hyperbolic cotangent (coth x) is defined as the ratio between the hyperbolic cosine (cosh x) and the hyperbolic sine (sinh x): $$\operatorname{coth} x =\frac{\cosh x}{\sinh x}$$

Apply the Quotient Rule

To derive the formula, use the quotient rule for differentiation, which states that for any two differentiable functions f(x) and g(x), the derivative of \(\frac{f(x)}{g(x)}\) is given by: $$\frac{d}{d x}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$$ In this case, we have f(x) = cosh x and g(x) = sinh x. We are given the respective derivatives in the exercise: \(\frac{d}{d x}(\cosh x) = \sinh x\) and \(\frac{d}{d x}(\sinh x) = \cosh x\). Thus, we can apply the quotient rule directly:

Substitute f(x) and g(x)

Substituting f(x) and g(x) with their respective derivatives, we get: $$\frac{d}{d x}(\operatorname{coth} x) = \frac{\sinh x \cdot \sinh x - \cosh x \cdot \cosh x}{(\sinh x)^2}$$

Simplify the expression

To find the desired derivative formula, we can simplify the expression obtained above: $$\frac{d}{d x}(\operatorname{coth} x) = \frac{\sinh^2 x - \cosh^2 x}{\sinh^2 x}$$

Use the hyperbolic identity

We can now use the identity \(\cosh^2 x - \sinh^2 x = 1\) to rewrite the numerator: $$\frac{d}{d x}(\operatorname{coth} x) = \frac{-1}{\sinh^2 x}$$ Finally, we can write the derivative using the square of the hyperbolic cosecant function:

Express in terms of csch^2 x

We know that \(\operatorname{csch} x = \frac{1}{\sinh x}\), so \(\operatorname{csch}^{2} x = \frac{1}{\sinh^2 x}\). Replacing the denominator with the hyperbolic cosecant function, we get the final result: $$\frac{d}{d x}(\operatorname{coth} x) = - \operatorname{csch}^2 x$$

Key concepts

Derivative Formulas

When dealing with hyperbolic functions, knowing the derivative formulas is essential. These are similar to those for trigonometric functions but with subtle differences. For example, the derivative of the hyperbolic cosine (\(\cosh x\)) is the hyperbolic sine (\(\sinh x\)) and vice versa. Here's the intuition:

• The derivative of \(\cosh x\) with respect to \(x\) is \(\sinh x\).
• The derivative of \(\sinh x\) with respect to \(x\) is \(\cosh x\).

This derivative relationship underlies many computations involving hyperbolic functions. For example, when finding the derivative of a more complex function, like the hyperbolic cotangent \(\operatorname{coth} x\), these foundational derivatives are utilized repeatedly. Recognizing these basic derivatives is the initial step in tackling the derivative of composite or inverse hyperbolic functions.

Quotient Rule

The quotient rule is a helpful technique in calculus for differentiating functions expressed as a quotient of two functions. This rule becomes particularly useful when working with hyperbolic functions like \(\operatorname{coth} x\). The rule states:

• If \(f(x)\) and \(g(x)\) are differentiable, then the derivative of \(\frac{f(x)}{g(x)}\) is \(\frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\).

In our example, \(\operatorname{coth} x = \frac{\cosh x}{\sinh x}\), with \(\cosh x\) and \(\sinh x\) as the numerator and denominator. By applying the quotient rule, we compute:\[\frac{d}{dx}\left(\operatorname{coth} x\right) = \frac{\sinh x \times \sinh x - \cosh x \times \cosh x}{(\sinh x)^2}\]This shows the application of the rule, step by step, illustrating how it simplifies the differentiation process. Understanding the quotient rule helps you manage more complicated derivations with multiple functions involved.

Hyperbolic Identities

In calculus, hyperbolic identities play a critical role similar to trigonometric identities. A key hyperbolic identity we used in our exercise is:

• \(\cosh^2 x - \sinh^2 x = 1\)

This identity, like the Pythagorean identity for trigonometric functions, helps simplify expressions. For instance, in differentiating \(\operatorname{coth} x\), the expression \(\sinh^2 x - \cosh^2 x\) was transformed into \(-1\) using this identity. This simplification is essential in reaching the final expression involving \(\operatorname{csch}^2 x\). By identifying and employing these identities, complex hyperbolic function calculations are more manageable and efficient, making it easier to derive precise results. Understanding these identities is crucial for anyone diving into deeper calculus involving hyperbolic functions.